SEBA Class 7 Mathematics Chapter 10 Practical Geometry Solutions English Medium, SEBA Class 7 Maths Notes in English Medium, SEBA Class 7 Mathematics Chapter 10 Practical Geometry Notes to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 7 Mathematics Chapter 10 Practical Geometry Solutions in English and select needs one.
SEBA Class 7 Mathematics Chapter 10 Practical Geometry
Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 7 General Maths Textual Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 7 Mathematics Chapter 10 Practical Geometry Solutions for All Subject, You can practice these here.
Practical Geometry
Chapter – 10
PART – II
Exercise – 10.1
1. Draw a line m. Take a point ‘A’ which is not in the line m. Draw a line through ‘A’ which is parallel to m.
Ans:
Steps of Construction:
1. A line ‘m’ is drawn.
2. Outside the line ‘m’, Point A is taken.
3. A point ‘B’ is taken on the line ‘m’.
4. Points A and B are joined.
5. With the point B as the centre, draw an arc and this are cuts the line ‘m’ at C and AB at D.
6. With the point C as centre another arc having same radius is drawn and this arc cuts AB at the point E.
7. With the point E an another arc having radius equal to CD and this arc cuts EF at G.
8. Now, A and G are joined and line ‘n’ is drawn which is the required line parallel to the line ‘m’.
2. Draw a line segment AB of length 10 cm. Draw another ray AX at point A such that ∠BAX = 60°, Take a point D on AX so that AD = 4 cm. Draw a line through point D which the parallel to the AB.
Ans:
Steps of construction:
1. A line AB = 10 cm is drawn.
2. At the point A, a ray AX is drawn in such a way ∠BAX will be 60°. From the line AX, AD = 4cm part is cut out.
3. Now, at D, ∠EDA = 60° is drawn equal to the ∠BAX = 60°, ED is extended to C. CE line will be parallel to AB.
3. Draw a line PQ. Draw a perpendicular line at any point of PQ. Take a point R on the perpendicular line which is 5-6 cm away from PQ. Draw a line through R which is parallel to the PQ.
Ans:
Steps of Construction:
1. PQ line is drawn.
2. A point ‘S’ is taken on the line PQ. Now, at the point S, ST perpendicular is drawn.
3. A point ‘R’ is taken at a distance of 5.6 cm from ‘S’.
4. Through the point ‘R’, a line segment ‘m’ is drawn in such a way that the line is perpendicular to ST. Now, the line ‘m’ will be parallel to PQ.
4. Draw a line segment XY of 7-5 cm. Draw two perpendicular lines of X and Y. Identify two points A and B in the perpendicular line which are 5 cm away from the XY. Through point A draw a line parallel to XY. Will this line pass through the point B?
Ans:
Steps of Construction:
1. A line XY measuring 7-5 cm is drawn.
2. At the points X and Y, two perpendiculars PX and QY are drawn.
3. On the two perpendiculars PX and QY, two points A and B are taken at a distance of 5 cm from the line XY.
4. Now, through the point A, EF line is drawn parallel to XY which passes through the point B.
| Exercise – 10.2 |
1. Construct triangle ABC so taht–
(i) AB = 3 cm
BC = 4 cm
AC = 2.5 cm
Ans:
Let ABC be a triangle of which AB = 3 cm, BC = 4 cm and AC = 2.5 cm.
Step1: Draw a line BC of length 4 cm.
Step2: Considering B as centre an arc of radius 3 cm is drawn.
Step3: Again, Considering C as centre an arc of radius 2.5 is drawn. This arc intersects each other at A.
Step4: Now, A, B and A, C are joined. In this way the triangle ABC is constructed of which AB = 3 cm, BC = 4 cm and AC = 2.5 cm
(ii) AB = 6 cm
BC = 4 cm
AC = 7 cm
Ans:
Let AB = 6 cm, BC = 4 cm, AC = 7cm of a ∆ABC.
Step1: Draw a line AC of length 7 cm.
Step2: Considering A as centre an arc of radius 6 cm is drawn.
Step3: Again, considering C as centre an arc of radius 4 cm is drawn. These two arc intersect each other at B.
Step4: BA and BC are joined.
Hence ∆ABC is a required triangle.
2. Construct an isosceles triangle base of which is 5 cm and two equal sides are of 4 cm.
Ans:
Steps of constructions:
Step1: Draw a line AB of length 5 cm.
Step2: Considering A as centre is drawn an arc of radius 4 cm. and again considering B as centre is drawn another arc of same radius. These two arc are intersected at C.
Step4: CA and CB are joined.
Hence ∆ABC is a required triangle.
3. Construct an equilateral ∆DEF of which DE = 6.5 cm.
Ans:
Steps of construction:
Step1: A line DE of length 6.5 cm is drawn.
Step2: Considering D as centre, an arc of radius 6.5 cm is drawn and another arc of same radius, considering E as the centre is drawn. These two arc are intersected at the point F.
Step3: FD and FE are joined.
Hence ∆FDE is the required triangle.
4. Construct the ∆XYZ of which XY = YZ = 5.5 cm and XZ = 4 cm.
Ans:
The triangle XYZ of which XY = YZ = 5.5 cm and XZ = 4 cm.
Steps of construction:
Step1: A line XZ of length 4 cm is drawn.
Step2: Considering X and Z as centres, two arcs each of radius 5.5 cm are drawn.
Step3: Two arcs are intersected each other at Y.
Step4: YX and YZ are joined.
Step5: Therefore, ∆XYZ is the required isosceles triangle.
| Exercise – 10.3 |
1. Construct a ∆ABC of which AB = 6 cm, BC = 7 cm and ∠B = 75°.
Ans:
Let ∆ABC is a triangle of which AB = 6 cm, BC = 7 cm and ∠B = 75°. It is required to construct a triangle.
Steps of construction:
Step1: BC a line is drawn of length 7 cm.
Step2: A + B, ∠XBC = 75° is constructed.
Step3: Considering B as centre an arc of radius AB = 6 cm. is drawn which cuts BX at A.
Step4: Now, AB has joined. Hence, ∆ABC is the required triangle.
2. Construct a ∆PQR of which QR = PR = 8 cm and ∠R = 60°.
Ans:
Steps of Construction:
Step1: QR a line is drawn of length 8 cm.
Step2: At the point R, RX is drawn in such a way it makes an angle measuring 60° with QR.
Step3: Considering R as centre an arc of radius 8 cm is drawn which cuts RX at P.
Step4: Join PR. Hence, ∆PQR is the required triangle.
3. Construct a ∆DEF of which DE = 5.5cm , DF = 7.5cm and angle D = 80°.
Ans:
A triangle DEF of which DE = 5.5 cm., DF = 7.5 cm and ∠D = 80°. It is required to construct a triangle.
Steps of construction:
Step1: DF a line segment is drawn of length 7-5 cm.
Step2: At the point D, DX line is drawn in such a way that it makes ∠80° with DF.
Step3: Considering D as centre an arc of radius 5.5 cm is drawn which cuts DX at E.
Step4: EF is joined. Hence ∆DEF is the required triangle.
4. Construct a right angle triangle, the two perpendicular lines each of which is 4.5 cm.
Ans:
Steps of Construction:
Step1: A line segment QR = 4.5 cm is drawn.
Step2: At the point Q, QX 丄 QR is drawn.
Step3: Considering R as centre an arc of radius 4.5 cm is drawn which cuts QX at P.
Step4: PR is joined. Hence ∆PQR is the required triangle.
5. Construct a triangle of which two sides are 5.5 cm and 6.5 cm and the angle between two sides is 120°.
Ans:
Steps of Construction:
Step1: A line segment BC = 6.5 cm is drawn.
Step 2: Considering B as centre BX is drawn in such a way that it makes 120° with CB.
Step3: Considering B as centre an arc of radius 5.5 cm is drawn which cuts BX at A.
Step4: AC is joined. Hence ∆ABC is the required triangle.
| Exercise – 10.4 |
1. Construct a ∆ABC of which ∠B = 65°, ∠C = 55°and BC = 7cm.
Ans:
∆ΑBC of which ∠B = 65°, ∠C = 55° and BC = 7 cm. It is required to construct a triangle.
Steps of Construction:
Step1: A line segment BC = 7cm is drawn.
Step2: At B, ∠CBX = 65°and at C, ∠BCY = 55° are drawn. BX and CY rays are intersected at A. Finally, ∆ABC is constructed of which ∠B = 65°, ∠C = 55° and BC = 7cm.
2. Construct a triangle of which one side is 6.5 cm and the angles formed at two ends of the side are 45° and 75°.
Ans:
Steps of Construction:
Step1: A line AB = 6.5 cm is drawn.
Step2: At the point A, AP line is drawn in such a way that an angle of 45° with AB.
Step3: Again, at the point B, BQ line is drawn in such a way that an angle of 75° with BA.
Step4: Two lines AP and BQ intersect at C. Finally, ∆ABC is formed.
3. Construct a ∆PQR of which ∠P = 60°, ∠Q = 50° and QR = 6 cm.
Ans:
Steps of Construction:
Step1: A line segment QR = 6 cm is drawn.
Step2: At the point Q, QX is drawn in such a way, an angle of 50° is formed with QP.
Step3: At the point R, RY is drawn in such a way, an angle of 70° is formed with QR.
According to the property of sum of the angles of a triangle-
m ∠Q + m∠P = m ∠R = 180°
⇒ 50° + 60° + m ∠R = 180°
⇒ m ∠R = 180°- 110° = 70°
Step4: Two rays are intersected at P and finally ∆PQR is constructed.
4. Construct a ∆XYZ of which XZ = 5.5cm ∠X = 100° and ∠Z = 30°.
Ans:
Steps of Construction:
Step1: A line segment XZ = 5.5 cm is drawn.
Step2: At the point X, XP ray is drawn in such a way that an angle of 100° with XZ.
Step3: Again, at the point Z, ZR ray is drawn in such a way an angle of 30° is formed with XZ.
Step4: Two rays XP and ZR are intersected of Y and finally ∆XYZ is constructed.
| Exercise – 10.5 |
1. Construct ∆ABC of which ∠A = 90°, BC = 8 cm and AB = 5cm.
Ans:
Steps of Construction:
Step1: A line segment AB = 5 cm is drawn.
Step2: At the point A, AX 丄 AB is drawn.
Step3: Considering B as centre an arc of radius 8 cm is drawn.
Step4: This arc cuts AX at C.
Step5: CB is joined and finally a ∆ABC is constructed.
2. The length of hypotenuse of a right-angled triangle is 10 cm and another side is 6 cm. Construct the triangle.
Ans:
Steps of Construction:
Step1: A line segment QR = 6 cm is drawn.
Step2: At Q, QX 丄 QR is drawn.
Step3: Considering R as centre an arc of radius 10 cm is drawn.
Step4: This arc cuts QX at P.
Step5: PR is joined and finally ∆PQR is constructed.
3. Construct ∆PQR of which ∠Q = 90°, PR = 7.5 cm and PQ = 5 cm
Ans:
Steps of Construction:
Step1: A line segment PQ = 5 cm is drawn.
Step2: At the point Q, angle ∠POX = 90° is drawn.
Step3: Considering Q as centre an arc of radius 7.5 cm is drawn. This arc cuts QX at R.
Step4: R, P are joined. This is the ∆PQR constructed.
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